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InterviewSolution
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If the Mach number of the moving shock wave relative to the laboratory Ms = 1.25 then what is the Mach number of the reflected shock wave in a calorically perfect gas?(a) 1.2331(b) 1.4821(c) 1.5671(d) 1.1102This question was posed to me in an online interview.This key question is from Reflected Shock Wave in section Unsteady Wave Motion of Aerodynamics |
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Answer» CORRECT choice is (a) 1.2331 Easy explanation: Given, Ms = 1.25 The relation between Ms and MR is: \(\frac {M_R}{M_R ^2 – 1} = \frac {M_s}{M_s ^2 – 1} \sqrt {1 + \frac {2(γ – 1)}{(γ + 1)^2}(M_s^2 – 1)\big (γ + \frac {1}{M_s ^2} \big )}\) Substituting the values we get \(\frac {M_R}{M_R ^2 – 1} = \frac {1.25}{1.25^2 – 1}\sqrt {1 + \frac {2(1.4 – 1)}{(1.4 + 1)^2}(1.25^2 – 1)\big (1.4 + \frac {1}{1.25^2} \big )} \) \(\frac {M_R}{M_R ^2 – 1} = \frac {1.25}{1.25^2 – 1}\sqrt {1 + \frac {2(1.4 – 1)}{(1.4 + 1)^2}(1.25^2 – 1)\big (1.4 + \frac {1}{1.25^2} \big )} \) \(\frac {M_R}{M_R ^2 – 1}\) = 2.20\(\sqrt {1 + \frac {0.8}{5.76}(0.5625)(2.04)}\) = 2.3689 MR = 2.3689 MR^2 – 2.3689MR On solving the QUADRATIC equation we get two RESULTS: MR = 1.2331, – 0.81 Negative values of Mach NUMBER is not POSSIBLE, hence MR = 1.2331 |
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